Comments on: Computing argmax fast in Python

By: Protector one

Protector one — Wed, 07 May 2008 20:31:05 +0000

I’m sorry SwiftCoder, but I believe that is a max. It is equivalent to:

m = max(array)

My (arg)max:

lambda array: max([array[i],i] for i in range(len(array)))[1]

Cheers!

By: SwiftCoder

SwiftCoder — Tue, 29 Jan 2008 14:20:38 +0000

You should also try a functional idiom - typically better and faster:

array = [1.0, 2.0, 3.0, 1.0]

m = reduce(max, array)

By: Anonymous

Anonymous — Sun, 13 Feb 2005 14:16:10 +0000

You can try this:

def positionMax(sequence, excludedIndexesSet=None):
if not badindexes:
return max( izip(sequence, xrange(len(sequence)) ) )[1]
else:
badindexes = set(badindexes)
return max( (e,i) for i,e in enumerate(sequence) if i not in badindexes )[1]